SOLUTION: Let tan(theta)=3 with theta om Quadrant I and find the following.
sec2(theta)
This is what I have done
found tan(theta)=3/1 cos(theta)=sqrt10/10 sin(theta)=3*sqrt10/10
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-> SOLUTION: Let tan(theta)=3 with theta om Quadrant I and find the following.
sec2(theta)
This is what I have done
found tan(theta)=3/1 cos(theta)=sqrt10/10 sin(theta)=3*sqrt10/10
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Question 1054131: Let tan(theta)=3 with theta om Quadrant I and find the following.
sec2(theta)
This is what I have done
found tan(theta)=3/1 cos(theta)=sqrt10/10 sin(theta)=3*sqrt10/10
from what I was taught I use the double angle formula cos2(theta)=cos^2(theta)-sin^2(theta), plug in your cos and sin and then sec2(theta) would be 1/cos2(theta)
I keep trying to find the cos2(theta) value but keep coming up with the wrong answer after plugging it in to sec2(theta) (checked in the back of my book) Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Let tan(theta)=3 with theta om Quadrant I and find the following.
sec2(theta)
This is what I have done
found tan(theta)=3/1 cos(theta)=sqrt10/10 sin(theta)=3*sqrt10/10
from what I was taught I use the double angle formula cos2(theta)=cos^2(theta)-sin^2(theta), plug in your cos and sin and then sec2(theta) would be 1/cos2(theta)
I keep trying to find the cos2(theta) value but keep coming up with the wrong answer after plugging it in to sec2(theta) (checked in the back of my book)
-------------------
tan(theta)=3
tan = y/x = 3/1
r = sqrt(3^2 + 1^2) = sqrt(10)
---
cos(t) = x/r = 1/sqrt(10)
cos(2t) = 2cos^2(t) - 1 = (1/5) - 1 = -4/5
sec(2t) = -5/4
===========================
Using the sine as you did:
tan(theta)=3
tan = y/x = 3/1
r = sqrt(3^2 + 1^2) = sqrt(10)
sin(t) = y/r = 3sqrt(10)/10
---
cos(2t) = cos^2(t) - sin^2(t) = (1/10) - 9/10 = -4/5
sec(2t) = -5/4